Divergence of a function involving delta distribution

Question

Given the vector field $\vec{b}$ defined in this way: $$ \vec{b}=-4\pi g\int_0^{\infty} dt \frac{d\vec{s}}{dt}\delta^3(\vec{r}-\vec{s(t)}),$$ where $\vec{s(t)}: t\in[0,+\infty)\rightarrow \mathbb{R^3}$ is the parametrization in respect to arc-length of a simple curve with $\vec{s(0)}=0$, I have to show that $$\nabla \cdot \vec{b}=4\pi g\delta^3(\vec{r}).$$ I've encountered this problem in a book of Physics, which is "Quantum mechanics: a new introduction" by K. Konishi and G.Paffuti, in a discussion on page 423 related to the Dirac string. However after many trials I'm wondering if the expression for $\vec{b}$ does make sense. Because as it is I don't think that it is possible to claim the previous equation. What do you think?

Answer 1

Assuming $\|\vec{s}(t)\| \to \infty$ as $t \to \infty$, I can derive the equality with signs reversed:

In mathematics, Dirac delta and related "functions" are realized as distribution (a.k.a. generalized functions). Intuitively, a distribution is a quantity defined across the space that can only be felt by a "probing needle". Formally, a distribution is a continuous linear functional acting on a suitable space of test functions.

So, let $\varphi \in C_c^{\infty}(\mathbb{R}^3)$ be any compactly supported smooth function on $\mathbb{R}^3$. Then, regarding $\vec{b}$ as a distribution in variable $\vec{r}$, its distributional derivative $\nabla \cdot \vec{b}$ is specified by the pairing

$$ \langle \nabla \cdot \vec{b}, \varphi \rangle = \int_{\mathbb{R}^3} \mathrm{d}^3\vec{r} \, [(\nabla \cdot \vec{b})\varphi(\vec{r})] = -\int_{\mathbb{R}^3} \mathrm{d}^3\vec{r} \, [\vec{b} \cdot \nabla \varphi(\vec{r}) ] = - \langle \vec{b}, \nabla \varphi \rangle, $$

where we invoked "integration by parts" in the second step. (In reality, this is the defining equation for the distribution $\nabla \cdot \vec{b}$.) Now plugging the definition of $\vec{b}$,

\begin{align*} \langle \nabla \cdot \vec{b}, \varphi \rangle &= 4\pi g \int_{\mathbb{R}^3} \mathrm{d}^3\vec{r} \int_{0}^{\infty} \mathrm{d}t \, \frac{\mathrm{d}\vec{s}(t)}{\mathrm{d}t} \delta^3(r - s(t)) \cdot \nabla \varphi(\vec{r})\\ &= 4\pi g \int_{0}^{\infty} \mathrm{d}t \, \frac{\mathrm{d}\vec{s}(t)}{\mathrm{d}t} \cdot \nabla \varphi(\vec{s}(t)) \\ &= 4\pi g \lim_{T \to \infty} \int_{0}^{T} \mathrm{d}t \, \frac{\mathrm{d}\vec{s}(t)}{\mathrm{d}t} \cdot \nabla \varphi(\vec{s}(t)) \\ &= 4\pi g \lim_{T \to \infty} \left[ \varphi(\vec{s}(T)) - \varphi(0) \right] \\ &= -4\pi g \varphi(0) \\ &= \langle -4\pi g \delta^3(\cdot), \varphi \rangle. \end{align*}

Therefore we obtain

$$ \nabla \cdot \vec{b} = -4\pi g \delta^3(\vec{r}). $$